JEE Main 2022PhysicsElectromagnetic InductionEasyNumerical

JEE Main 2022 — Electromagnetic Induction Question with Solution

JEE Main 2022 (25 Jul Shift 2)

Question

Magnetic flux (in weber) in a closed circuit of resistance 20 Ω varies with time ts as ϕ=8t2-9t+5. The magnitude of the induced current at t=0.25 s will be _____ mA.

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Show full solutionCorrect answer: 250
Correct answer
250

Step-by-step explanation

Given that magnetic flux ϕ=8t2-9t+5

According to Faraday's law of electromagnetic induction

emf=-dϕdt=-16t-9

At t=0.25 s

emf=-16×0.25-9=5 V

Current =emf  Resistance =5 V20 Ω

=14 A=10004 mA=250 mA

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About this question

This is a previous-year question from JEE Main 2022, covering the Electromagnetic Induction chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.